Help me to solve this calculus I question.. click on link.. Show the method ..

https://docs.google.com/document/pub?id=1enrHRY8cOT5Phj1QrnlZvXPC-gscGcB6Mo83Z1zZhBo

f(x)= (2x^2+3x)/e^x

A) no vert. Asymp. ( denom cannot = zero)
Horiz.asymp: use l'hopitals
Lim(x->inf) = inf/inf
= lim( 4x+3)/e^x -> inf/inf
= lim(4/e^x)
= 0
y= 0

B) to find inc.and dec, find y' by quotient rule.

Y' =[e^x(4x+3)- (2x^2+3x)*e^x]/ [e^(2x)]

= (-2x^2+x+3)/e^x=0

Solve -2x^2+x+3=0

-(2x-3)(x+1)=0

x=3/2 or -1

------(-1)-----(3/2)----- test the sign of y'
(-)............(+)...........(-)

Decreasing on (-inf, -1) and (3/2, inf)
Increasing on (-1,3/2)

C) min at (-1,? ); max at (3/2,?) find y values from the original function.

D) use the second deriv. To test for inflection points and concavity.

Y" = (2x^2-5x-2)/e^x by quotient rule.

Set equal to zero, solve by quadratic formula.

X=[ 5+/-sqr(41)]/4
x= -.35 or 2.85

Test for concavity

--------(-.35)-----(2.85)-----
(+).... .........(-).... .........(+)
up..............down.........up
(-inf, -.35). (-.35,2.85).....(2.85, inf)

Infl points at x= -.35 and 2.85

Find y values from the original function.

Double check all values on your calculator.

Hoping this helps!

47 R

Not so much